Lectures on The Riemann Zeta–Function

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Lectures on

The Riemann Zeta–Function

By

K. Chandrasekharan

Tata Institute of Fundamental Research, Bombay 1953

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Lectures on the Riemann Zeta-Function

By

K. Chandrasekharan

Tata Institute of Fundamental Research 1953

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The aim of these lectures is to provide an intorduc- tion to the theory of the Riemann Zeta-function for stu- dents who might later want to do research on the subject.

The Prime Number Theorem, Hardy’s theorem on the Zeros ofζ(s), and Hamburger’s theorem are the princi- pal results proved here. The exposition is self-contained, and required a preliminary knowledge of only the ele- ments of function theory.

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6 Some estimates forζ(s) . . . 97 7 Functional Equation (Third Method) . . . 99 13 The Zeta Function of Riemann (Contd) 105 8 The zeros ofζ(s) . . . 105 14 The Zeta Function of Riemann (Contd) 113 9 Riemann-Von Magoldt Formula . . . 113 15 The Zeta Function of Riemann (Contd) 121 10 Hardy’s Theorem . . . 121 16 The Zeta Function of Riemann (Contd) 127 17 The Zeta Function of Riemann (Contd) 135 12 The Prime Number Theorem . . . 135 13 Prime Number Theorem and the zeros ofζ(s) . . . 145 14 Prime Number Theorem and the magnitude of pn . . . . 146

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Lecture 1

The Maximum Principle

Theorem 1. If D is a domain bounded by a contour C for whichCauchy’s 1

theorem is valid, and f is continuous on C regular in D, then “|f| ≤M on C” implies “|f| ≤ M in D”, and if|f|= M in D, then f is a constant.

Proof. (a) Let z_o∈D, n a positive integer. Then

|f (zo)|ⁿ=

1 2πi

Z

C

{f (z)}ⁿdz z−z_o

≤ l_c·Mⁿ 2πδ ,

where lc =length of C,δ=distance of zofrom C. As n→ ∞

|f (z)| ≤M.

(b) If|f (zo)|= M, then f is a constant. For, applying Cauchy’s inte- gral formula to d

dz

{f (z)}ⁿ

, we get

|n{f (zo)}ⁿ⁻¹· f^′(zo)|=

1 2πi

Z

C

fⁿdz (z−zo)²

≤ l_CMⁿ 2πδ² , 1

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2 1. The Maximum Principle so that

|f^′(zo)| ≤ lcM 2πδ² · 1

n →0, as n→ ∞ Hence

|f^′(zo)|=0.

(c) If|f (zo)|= M, and|f^′(zo)|=0, then f^′′(zo)=0, for

2

d² dz²

{f (z)}ⁿ

=n(n−1){f (z)}ⁿ⁻²{f^′(z)}²+n{f (z)}ⁿ⁻¹f^′′(z).

At zowe have d² dz²

{f (z)}ⁿ

z=zo =n fⁿ⁻¹(Z0) f^′′(z0), so that

|nMⁿ⁻¹f^′′(z0)|=

2!

2πi Z

C

{f (z)}ⁿdz (z−z_o)³

≤ 2!lc

2πδ³Mⁿ,

and letting n→ ∞, we see that f^′′(zo)=0. By a similar reasoning we prove that all derivatives of f vanish at z₀ (an arbitrary point of D). Thus f is a constant.

Remark . The above proof is due to Landau [12, p.105]. We shall now show that the restrictions on the nature of the boundary C postulated in the above theorem cab be dispensed with.

Theorem 2. If f is regular in a domain D and is not a constant, and M=max

z∈D |f|, then

|f (zo)|< M, zo∈D.

For the proof of this theorem we need a

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1. The Maximum Principle 3 Lemma. If f is regular in|z−z0| ≤r, r>0, then

|f (z0)| ≤Mr, where Mr = max

|z−zo|=r|f (z)|, and|f (zo)| = Mr only if f (z) is constant for

3

|z−zo|=r.

Proof. On using Cauchy’s integral formula, we get f (zo)= 1

2πi Z

|z−zo|=n

f (z) z−z0

dz

= 1 2π

Z2π

0

f (z_o+re^iθ)dθ. (1) Hence

|f (zo)| ≤Mr.

Further, if there is a point ζ (such that) |ζ −zo| = r, and|f (ζ)| <

Mr, then by continuity, there exists a neighbourhood ofζ, on the circle

|z − z_o| = r, in which |f (z)| ≤ M_r − ε, ε > 0 and we should have

|f (z_o)|<M_r; so that “|f (z_o)|= M_r” implies “|f (z)|= M_reverywhere on

|z−zo|=r”. That is|f (zo+reîθ)|=|f (zo)|for 0≤θ≤2π, or f (zo+reîθ)= f (zo)eîϕ, 0≤ϕ≤2π On substituting this in (1), we get

1= 1 2π

Z2π

0

cosϕdθ

Sinceϕis a continuous function ofθand cosϕ≤ 1, we get cosϕ·1 for

allθ, i.e.ϕ=0, hence f (s) is a constant.

Remarks . The Lemma proves the maximum principle in the case of a 4

circular domain. An alternative proof of the lemma is given below [7, Bd 1, p.117].

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4 1. The Maximum Principle Aliter. If f (z0+re^iθ)=φ(θ) (complex valued) then

f (z₀)= 1 2π

Z2π

0

φ(θ)dθ

Now, if a and b are two complex numbers, |a| ≤ M, |b| ≤ M and a, b, then|a+b|< 2M. Hence ifφ(θ) is not constant for 0 ≤θ <2π, then there exist two pointsθ₁,θ₂such that

|φ(θ1)+φ(θ2)|=2Mr−2ε, say, whereε >0 On the other hand, by regularity,

|φ(θ1+t)−φ(θ1)|< ε/2,for 0<t< δ

|φ(θ2+t)−φ(θ2)|< ε/2,for 0<t< δ











Hence

|φ(θ₁+t)+φ(θ₂+t)|<2M_r−ε, for 0<t< δ.

Therefore

2π

Z

0

φ(θ)dθ=

θ1

Z

0

+

θ1+δ

Z

θ1

+

θ2

Z

θ1+δ

+

θ2+δ

Z

θ2

+

2π

Z

θ2+δ

=







θ1

Z

0

+

θ2

Z

θ1+δ

+ Z2π

θ2+δ







φ(θ)dθ+ Zδ

0

[φ(θ₁+t)+φ(θ₂+t)] dt

Hence

|

2π

Z

0

φ(θ) dθ| ≤ Mr(2π−2δ)+(2Mr−ε)δ=2πMr−εδ

| 1 2π

Z2π

0

φ(θ)dθ|< M_r

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1. The Maximum Principle 5 Proof of Theorem 2. Let zo ∈ D. Consider the set G1of points z such

5

that f (z) , f (z_o). This set is not empty, since f is non-constant. It is a proper subset of D, since zo ∈ D, zo <G1. It is open, because f is continuous in D. Now D contains at least one boundary point of G1. For if it did not, G^′₁∩D would be open too, and D= (G₁∪G^′₁)D would be disconnected. Let z1be the boundary point of G1such that z1∈D. Then z1 < G1, since G1 is open. Therefore f (z1) = f (zo). Since z1 ∈ BdG1

and z₁∈D, we can choose a point z₂ ∈G₁such that the neighbourhood of z1defined by

|z−z1| ≤ |z2−z1|

lies entirely in D. However, f (z₂) , f (z₁), since f (z₂) , f (z_o), and f (zo) = f (z1). Therefore f (z) is not constant on |z − z1| = r, (see (1) on page 3) for, if it were, then f (z₂) = f (z₁). Hence if M^′ =

|z−zmax1|=|z2−z1||f (z)|, then

|f (z₁)|< M^′ ≤ M=max

z∈D |f (z)| i.e. |f (zo)|< M

Remarks. The above proof of Theorem 2 [7, Bd I, p.134] does not make use of the principle of analytic continuation which will of course provide an immediate alternative proof once the Lemma is established.

Theorem 3. If f is regular in a bounded domain D, and continuous in ¯D, then f attains its maximum at some point in Bd D, unless f is a constant.

Since D is bounded, ¯D is also bounded. And a continuous function on a compact set attains its maximum, and by Theorem 2 this maximum 6

cannot be attained at an interior point of D. Note that the continuity of

|f|is used.

Theorem 4. If f is regular in a bounded domain D, is non-constant, and if for every sequence{zn}, zn∈D, which converges to a pointξ∈Bd D, we have

lim sup

n→∞|f (z_n)| ≤ M, then|f (z)|< M for all z∈D. [17, p.111]

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6 1. The Maximum Principle Proof. D is an Fσ, for define sets Cn by the property: Cn consists of all z such that|z| ≤ n and such that there exists an open circular neigh- bourhood of radius 1/n with centre z, which is properly contained in D.

Then Cn ⊂C_n+1, n=1,2, . . .; and Cnis compact.

Define

maxZ∈Cn|f (z)| ≡mn.

By Theorem 2, there exists a z_n ∈ Bd C_n such that|f (z_n)| = m_n. The sequence{m_n}is monotone increasing, by the previous results; and the sequence {zn} is bounded, so that a subsequence {znp} converges to a limitξ∈Bd D. Hence

lim|f (znp)| ≤M i.e. lim mnp ≤ M or mn< M for all n i.e. |f (z)|< M. z∈D.

N.B. Thatξ ∈ Bd D can be seen as follows. Ifξ ∈ D, then there exists an Nosuch thatξ∈CNo ⊂D, and|f (ξ)|<mNo ≤lim mn, whereas

7

we have f (z_n_p)→ f (ξ), so that|f (z_n_p)| → |f (ξ)|, or lim m_n=|f (ξ)|. Corollaries. (1) If f is regular in a bounded domain D and contin-

uous in D, then by considering e^{f (z)} and e⁻^{i f (z)} instead of f (z), it can be seen that the real and imaginary parts of f attain their maxima on Bd D.

(2) If f is an entire function different from a constant, then M(r)≡l.u.b

|z|=r |f (z)|, and A(r)≡l.u.b

|z|=r Re{f (z)}

are strictly increasing functions of r. Since f (z) is bounded if M(r) is, for a non-constant function f , M(r)→ ∞with r.

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1. The Maximum Principle 7 (3) If f is regular in D, and f , 0 in D, then f cannot attain a

minimum in D.

Further if f is regular in D and continuous in ¯D, and f ,0 in ¯D, and m = min

z∈Bd D|f (z)|, then|f (z)| > m for all z ∈ D unless f is a constant. [We see that m>0 since f ,0, and apply the maximum principle to 1/f ].

(4) If f is regular and , 0 in D and continuous in ¯D, and |f| is a constantγon Bd D, then f =γ in D. For, obviouslyγ , 0, and, by Corollary 3,|f (z)| ≥γfor z∈D, se that f attains its maximum in D, and hence f =γ.

(5) If f is regular in D and continuous in ¯D, and|f|is a constant on 8

Bd D, then f is either a constant or has a zero in D.

(6) Definition: If f is regular in D, thenαis said to be a boundary value of f atξ ∈ Bd D, if for every sequence z_n → ξ ∈ Bd D, zn∈D, we have

n→∞lim f (zn)=α

(a) It follows from Theorem 4 that if f is regular and non - con- stant in D, and has boundary-values{α}, and|α| ≤ H for eachα, then|f (z)|< H, z∈D. Further, if M^′ =max

α∈b.v|α|then M^′= M≡max

z∈D |f (z)|

For, |α|is a boundary value of |f|, so that |α| ≤ M, hence M^′ ≤ M. On the other hand, there exists a sequence{zn}, zn ∈ D, such that |f (zn)| → M. We may suppose that z_n → z_o (for, in any case, there exists a subsequence {z_n_p} with the limit zo, say, and one can then operate with the subsequence). Now zo < D, for otherwise by continuity

f (z_o) = lim

n→∞f (z_n) = M, which contradicts the maximum principle. Hence zo ∈ Bd D, and M is the boundary-value of|f|at zo. Therefore M≤ M^′, hence M= M^′.

(b) Let f be regular and non-constant in D, and have boundary- values{α} on Bd D. Let m = min

z∈D|f|; m^′ = min

α∈b.v.|α| > 0.

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8 1. The Maximum Principle Then, if f ,0 in D, by Corollary (6)a, we get

maxz∈D

1

|f (z)| = 1 m^′ = 1

m;

so that|f|>m=m^′in D. Thus if there exists a point z_o∈D

9

such that|f (zo)| ≤m^′, then f must have a zero in D, so that m^′ > 0 = m. We therefore conclude that, if m^′ > 0, then m^′ =m if and only if f ,0 in D [6, Bd. I, p.135].

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Lecture 2

The Phragmen-Lindelof principle

We shall first prove a crude version of the Phragmen-Lindelof theorem 10

and then obtain a refined variant of it. The results may be viewed as extensions of the maximum-modulus principle to infinite strips.

Theorem 1. [6, p.168] We suppose that

(i) f is regular in the stripα <x< β; f is continuous inα≤ x≤β (ii) |f| ≤ M on x=αand x=β

(iii) f is bounded inα≤ x≤β

Then,|f| ≤ M inα <x< β; and|f|= M inα <x< βonly if f is a constant.

Proof. (i) If f (x+iy) →O as y → ±∞uniformly in x,α≤ x ≤ β, then the proof is simple. Choose a rectangleα ≤ x ≤ β,|y| ≤ η with ηsufficiently large to imply |f|(x± iη) ≤ M for α ≤ x ≤ β. Then, by the maximum-modulus principle, for any z_o in the interior of the rectangle,

|f (zo)| ≤ M.

9

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10 2. The Phragmen-Lindelof principle (ii) If |f (x+iy)| 6→ 0, as y → ±∞, consider the modified function

f_n(z)= f (z)e^z²^/n. Now

|fn(z)| →0 as y→ ±∞uniformly in x, and

|fn(z)| ≤Me^c²^/n

on the boundary of the strip, for a suitable constant c.

Hence

11

|fn(zo)| ≤ Me^c²^/n

for an interior point zoof the rectangle, and letting n→ ∞, we get

|f (zo)| ≤ M.

If|f (z_o)| = M, then f is a constant. For, if f were not a constant, in the neighbourhood of zowe would have points z, by the maximum-modulus principle, such that|f (z)|> M, which is impossible.

Theorem 1 can be restated as:

Theorem 1^′: Suppose that

(i) f is continuous and bounded inα≤ x≤β (ii) |f (α+iy)| ≤ M1,|f (β+iy)| ≤M2for all y

Then

|f (xo+iyo)| ≤M₁^L(x^o⁾M^1−L(x₂ ^o⁾

forα <x_o< β,|y_o|<∞, where L(t) is a linear function of t which takes the value 1 atαand 0 atβ. If equality occurs, then

f (z)=cM₁^L(z)M₂¹⁻^L(z); |c|=1.

Proof. Consider

f₁(z)= f (z) M₁^L(z) M¹₂⁻^L(z) and apply Theorem 1 to f₁(z).

12

More generally we have [12, p.107]

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2. The Phragmen-Lindelof principle 11 Theorem 2. Suppose that

(i) f is regular in an open half-strip D defined by:

z= x+iy, α <x< β, y> η (ii) lim

z→ξ in D

|f| ≤ M, for ξ∈Bd D

(iii) f =O

exp

e^θπ^β−α^|^z^|

,θ <D uniformly in D.

Then,|f| ≤M in D; and|f|<M in D unless f is a constant.

Proof. Without loss of generality we can choose α=−π

2, β= +π/2, η=0 Set

g(z)= f (z).exp(−σe^−ikz)

≡ f (z).{ω(z)}^σ, say, whereσ >0,θ <k<1. Then, as y→+∞,

g=O

"

exp (

e^θ^|^z^|−σe^kycoskπ 2

#

uniformly in x, and 13

e^θ^|^z^|−σe^kycoskπ

2 ≤e^θ(y+π/2)−σe^kycoskπ 2

→ −∞, as y→ ∞, uniformly in x, sinceθ <k. Hence

|g| →0 uniformly as y→ ∞, so that

|g| ≤ M, for y>y^′, α < x< β.

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12 2. The Phragmen-Lindelof principle Let zo ∈D. We can then choose H so that

|g(z)| ≤M

for y=H,α <x< β, and we may also suppose that H >y0=im(zo).

Consider now the rectangle defined by α < x < β, y = 0, y = H.

Since|ω^σ| ≤1 we have

limz→ξ in D

|g| ≤ M

for every point ξ on the boundary of this rectangle. Hence, by the maximum-modulus principle,

|g(zo)|< M, unless g is a constant;

or |f (zo)|< M|e^σe^−ikzo| Lettingσ→0, we get

14

|f (z_o)|< M.

Remarks. The choice ofωin the above proof is suggested by the critical caseθ=1 of the theorem when the result is no longer true. For take

θ=1, α=−π/2, β=π/2, η=0, f =e^e⁻^iz Then

|eê^−iz|=e^Re(e^−iz⁾=e^Re^{ê^y−ix^}=eê^y^{cos x} So

|f|=1 on x=±π/2, and f =e^e^yon x=0.

Corollary 1. If f is regular in D, continuous on Bd D,|f| ≤ M on Bd D, and

f =O

e^e^|z|πθ^β⁻^α

, θ <1, then|f| ≤M in D.

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2. The Phragmen-Lindelof principle 13 Corollary 2. Suppose that hypotheses (i) and (iii) of Theorem 2 hold;

suppose further that f is continuous on Bd D,

f =O(y^a) on x=α, f =O(y^b) on x=β.Then f =O(y^c) on x=γ,

uniformly inγ, where c= pγ+q, and px+q is the linear function which equals a at x=αand b at x=β.

Proof. Letη >0. Defineϕ(z) = f (z)ψ(z), whereψis the single-valued 15

branch of (−zi)⁻^(pz+q)defined in D, and apply Theorem 2 toϕ. We first observe thatϕis regular in D and continuous in ¯D. Next

|ψ(z)|=|(y−ix)⁻^(px+q)⁻^ipy|

=|y−ix|^−(px+q)exp(py imlog y−ix

y )

=y⁻^(px+q)|1+O 1 y

!

|^O(1)exp

"

py (

−x

y +O 1 y²

!)#

=y⁻^(px+q)e⁻^px{1+O 1 y

! },

the O’s being uniform in the x^′s.

Thusϕ= O(1) on x =αand x= β. Sinceψ = O(y^k) = O(|z|^k), we see that condition (iii) of Theorem 2 holds forϕifθis suitably chosen.

Henceϕ=O(1) uniformly in the strip, which proves the corollary.

Theorem 3. [12, p.108] Suppose that conditions (i) and (iii) of Theorem 2 hold. Suppose further that f is continuous in Bd D, and

ylim→∞|f| ≤ M on x=α, x=β Then

y→∞lim|f| ≤ M uniformly in α≤ x≤β.

Proof. f is bounded in ¯D, by the previous theorem. Letη ≥ 0,P

> 0. 16

Let H=H(ε) be the ordinate beyond which|f|< M+εon x=α, x=β.

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14 2. The Phragmen-Lindelof principle Let h>0 be a constant. Then

z z+hi

<1 in the strip. Choose h=h(H) so large that

f (z)· z z+hi

< M+εon y=H (possible because

f· z

z+hi

≤ |f|< M). Then the function g(z)= f· z

z+hi

satisfies the conditions of Theorem 2 (with M+εin place of M) in the strip above y=H. Thus

|g| ≤M+εin this strip and so lim

y→∞|g| ≤ M+εuniformly.

That is,

lim|f| ≤ M+ε, since z

z+hi →1 uniformly in x. Hence lim|f| ≤ M.

Corollary 1. If conditions (i) and (iii) of Theorem 2 hold, if f is contin-

17

uous on Bd D, and if

lim|f| ≤











a on x=α, b on x=β, where a,0, b,0, then

ylim→∞|f| ≤e^px+q,

uniformly, p and q being so chosen that e^px+q=a for x=αand=b for x=β.

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2. The Phragmen-Lindelof principle 15

Proof. Apply Theorem 3 to g= f e⁻^(pz+q)

Corollary 2. If f =O(1) on x=α, f =o(1) on x=β, then f =o(1) on x=γ,α < γ≤β.

[if conditions (i) and (iii) of Theorem 2 are satisfied].

Proof. Take b=εin Corollary 1, and note that e^pγ+q →0 asε→0 for fixedγ, provided p and q are chosen as specified in Corollary 1.

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Lecture 3

Schwarz’s Lemma

Theorem 1. Let f be regular in|z|<1, f (o)=0. 18

If, for|z|<1, we have|f (z)| ≤1, then

|f (z)| ≤ |z|, for |z|<1.

Here, equality holds only if f (z)≡ c z and|c| =1. We further have

|f^′(o)| ≤1.

Proof. f (z)

z is regular for|z| < 1. Given o < r < 1 chooseρsuch that r < ρ < 1; then since|f (z)| ≤ 1 for|z| = ρ, it follows by the maximum modulus principle that

f (z) z

≤ 1

ρ,

also for|z| = r. Since the L.H.S is independent ofρ →1, we letρand obtain|f (z)| ≤ |z|, for|z|<1.

If for z₀, (|z₀| < 1), we have|f (z₀)| = |z₀|, then

f (z0) z₀

= 1, (by the maximum principle applied to f (z)

z ) hence f =cz,|c|=1.

Since f (z)= f^′(o)z+f^′′(o)z²+· · · in a neighbourhood of the origin, and since

f (z) z

≤ 1 in z 1, we get |f^′(o)| ≤ 1. More generally, we

have

17

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18 3. Schwarz’s Lemma Theorem 1^′: Let f be regular and|f| ≤ M in |z| < R, and f (o) = 0.

Then

|f (z)| ≤ M|z|

R , |z|<R.

In particular, this holds if f is regular in |z| < R, and continuous in

19

|z| ≤R, and|f| ≤M on|z|=R.

Theorem 2 (Caratheodory’s Inequality). [12, p.112] Suppose that (i) f is regular in|z|<R f is non-constant

(ii) f (o)=0

(iii) Re f ≤ ∪in|z|<R. (Thus∪>0, since f is not a constant).

Then,

|f| ≤ 2∪ρ

R−ρ, for|z|=ρ <R.

Proof. Consider the function

ω(z)= f (z) f (z)−2∪.

We have Re{f−2∪} ≤ −∪<0, so thatωis regular in|z|<R.

If f =u+iv, we get

|ω|= s

u²+v² (2∪ −u)²+v² so that

|ω| ≤1, since 2∪ −u≥ |u|. Butω(o)=o, since f (o)=o. Hence by Theorem 1,

|ω(z)| ≤ |z|

R, |z|<R.

But f =−2∪ω

1−ω. Now take|z|=ρ <R.

Then, we have,

20

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3. Schwarz’s Lemma 19

|f| ≤ 2∪ |ω|

1− |ω| ≤ 2∪ρ R−ρ N.B. If |f (zo)|= 2∪ρ

R−ρ, for|zo|<R,then f (z)= 2U·cz

1−cz, |c|=1.

More generally, we have

Theorem 2^′: Let f be regular in|z|<R; f non-constant f (o)=a_o=β+iγ

Re f ≤ ∪(|z|<R), so that∪ ≥β Then

|f (z)−ao| ≤ 2(∪ −β)ρ R−ρ for|z|=ρ <R

Remark . If f is a constant, then Theorem 2 is trivial. We shall now prove Borel’s inequality which is sharper than Theorem 2.

Theorem 3 (Borel’s Inequality). [12, p.114] Let f (z) = P^∞

n=0

a_nzⁿ, be regular in|z|<1. Let Re f ≤ ∪. f (o)=a_o=β+iγ.

Then

|an| ≤2(∪ −β)≡2 ∪1, say, n>0.

Proof. We shall first prove that|a1|<∪1, and then for general an. (i) Let f₁≡ P^∞

n=1

a_nzⁿ. Then Re f₁≤ ∪1. 21

For|z|=ρ <1, we have, by Theorem 2,

|f₁| ≤ 2∪1ρ 1−ρ i.e.

f1(z) z

≤ 2∪1

1−ρ

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20 3. Schwarz’s Lemma Now letting z→0, we get

|a1| ≤2∪1

(ii) Defineω=e^2πi/k, k a+ve integer.

Then

k

X

v=1

ω^νm=o











if m,0

& if m,a multiple of k

=k











if m=0

or if m= a multiple of k.

We have 1 k

k−1

X

r=0

f₁(ω^rz)= X∞ n=1

a_nkz^nk =g₁(z^k)≡g₁(ζ), say.

The series for g1is convergent for|z|<1, and so for|ζ|< 1. Hence g₁is regular for|ζ|<1. Since Re g₁≤ ∪1, we have

| coefficient ofζ| ≤ 2∪1 by the first part of the proof i.e. |ak| ≤

22

2∪1.

Corollary 1. Let f be regular in|z|<R, and Re{f (z)−f (o)} ≤ ∪1. Then

|f^′(z)| ≤ 2∪1R

(R−ρ)², |z|=ρ <R Proof. Suppose R=1. Then

|f^′(z)| ≤X

n|an|ρⁿ⁻¹≤2∪1

X nρⁿ⁻¹

This argument can be extended to the n^thderivative.

Corollary 2. If f is regular for every finite z, and e^{f (z)}=O(e^|z|^k), as|z| → ∞, then f (z) is a polynomial of degree≤k.

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3. Schwarz’s Lemma 21 Proof. Let f (z)=P

anzⁿ. For large R, and a fixedζ,|ζ|<1, we have Re{f (Rζ)}<cR^k.

By Theorem 3, we have

|anRⁿ| ≤2(2R^k−Re ao), n>0.

Letting R→ ∞, we get

an=0 if n>k.

Apropos Schwarz’s Lemma we give here a formula and an inequal-

ity which are useful.

Theorem 4. Let f be regular in|z−z_o| ≤r 23

f =P(r, θ)+Q(r, θ).

Then

f^′(zo)= 1 πr

2π

Z

o

P(r, θ)e⁻^iθdθ.

Proof. By Cauchy’s formula, (i) f (zo)= 1

2πi Z

|z−zo|=r

f (z)dz (z−z_o)² = 1

2πr

2π

Z

0

(P+iQ)e^−iθdθ By Cauchy’s theorem,

(ii) 0= 1 r² · 1

2πi Z

|z−zo|=r

f (z)dz= 1 2πr

2π

Z

0

(P+iQ)e^iθdθ We may change i to−i in this relation, and add (i) and (ii).

Then

f^′(zo)= 1 πr

2π

Z

0

P(r, θ)e⁻^iθdθ

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22 3. Schwarz’s Lemma Corollary. If f is regular, and|Re f ≤ M in|z−z0| ≤r, then

|f^′(zo)| ≤ 2M r .

Aliter: We have obtained a series of results each of which depended on the preceding. We can reverse this procedure, and state one general result from which the rest follow as consequences.

Theorem 5. [11, p.50] Let f (z)≡ P^∞

n=0

c_n(z−z₀)ⁿbe regular in|z−z₀|<R,

24

and Re f <∪. Then

|en| ≤ 2(∪ −Re co)

Rⁿ , n=1,2,3, . . . (5.1) and in|z−z_o| ≤ρ <R, we have

|f (z)− f (z0)| ≤ 2ρ

R−ρ{∪ −Re f (zo)} (5.2)

f⁽ⁿ⁾(z) n!

≤ 2R

(R−ρ)ⁿ⁺¹{∪ −Re f (z_o)}n=1,2,3, . . . (5.3) Proof. We may suppose z_o =0.

Set φ(z)=∪ − f (z)=∪ −c0− X∞

1

cnzⁿ

≡ X∞

0

bnzⁿ, |z|<R.

Letγdenote the circle|z|=ρ <R. Then bn= 1

2πi Z

γ

φ(z)dz zⁿ⁺¹ = 1

2πρⁿ

π

Z

−π

(P+iQ)e⁻^inθdθ,n≥0, (5.4) whereφ(r, θ)= P(r, θ)+iQ(r, θ). Now, if n≥1, thenφ(z)zⁿ⁻¹is regular isγ, so that

0= ρⁿ 2r

Zπ

−π

(P+iQ)e^inθdθ,n≥1 (5.5)

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3. Schwarz’s Lemma 23 Changing i to−i in (5.5) and adding this to (5.4) we get

bnρⁿ= 1 π

π

Z

−π

Pe⁻^nθidθ, n≥1.

But P=∪ −Re f ≥0 in|z|<R and so inγ. Hence, if n≥1, 25

|b_n|ρⁿ ≤ 1 π

Zπ

−π

|Pe⁻^nθi|dθ= 1 π

Zπ

−π

Pdθ=2 Re b_o, (5.6) on using (5.4) with n=0. Now lettingρ→R, we get

|bn|Rⁿ≤2βo ≡2 Re bo.

Since b_o=∪ −c_o, and b_n=−c_n, n≥1, we at once obtain (5.1). We then deduce, for|z| ≤ρ <R

|φ(z)−φ(o)|=| X∞

1

bnzⁿ| ≤ X∞

1

2β0

ρ R

n

= 2βoρ

R−ρ (5.7) and if n≥1,

|φ⁽ⁿ⁾(z)| ≤ X∞

r=n

r(r−1). . .(r−n+1)2β0ρ^r⁻ⁿ R^r

= d dρ

!n ∞

X

n=0

2β0

ρ R

r

= d dρ

!n

2β_eR R−ρ

=2 β_oR·n!

(R−ρ)ⁿ⁺¹ (5.8)

(5.7) and (5.8) yield the required results on substitutingφ = ∪ − f and

β₀=∪ −Re f (0).

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Lecture 4

Entire Functions

An entire function is an analytic function (in the complex plane) which 26

has no singularities except at∞. A polynomial is a simple example. A polynomial f (z) which has zeros at z₁, . . . ,z_ncan be factorized as

f (z)= f (0) 1− z z₁

!

. . . 1− z z_n

!

The analogy holds for entire functions in general. Before we prove this, we wish to observe that if G(z) is an entire function with no zeros it can be written in the form e^g(z)where g is entire. For consider G^′(z)

G(z); every (finite) point is an ‘ordinary point’ for this function, and so it is entire and equals g1(z), say. Then we get

log (G(z)

G(z0) )

=

z1

Z

z0

g₁(ζ)dζ =g(z)−g(z₀), say, so that

G(z)=G(z₀)e^g(z)−g(z⁰⁾=e^g(z)−g(z⁰^{)+log G(z}^o⁾

As a corollary we see that if G(z) is an entire function with n zeros, distinct or not, then

G(z)=(z−z₁). . .(z−z_n)e^g(z) 25

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26 4. Entire Functions where g is entire. We wish to uphold this in the case when G has an infinity of zeros.

Theorem 1 (Weierstrass). [15, p.246] Given a sequence of complex

27

numbers ansuch that

0<|a1| ≤ |a2| ≤. . .≤ |an| ≤. . .

whose sole limit point is∞, there exists an entire function with zeros at these points and these points only.

Proof. Consider the function 1− z

a_n

! e^Q^ν^(z),

whereQν(z) is a polynomial of degree q. This is an entire function which does not vanish except for z=an. Rewrite it as

1− z an

!

e^Q^ν^(z)=e^Q^ν^(z)+log

1−_an^z

=e⁻

z

an−_2a2^z2...+Qν(z)

, and choose

Qν(z)= z

an +. . .+ z^ν νa^νⁿ so that

1− z an

!

e^Q^ν^(z)=e⁻

_z

an

_ν+1

1 ν+1

...

≡1+∪n(z), say.

We wish to determineνin such a way that Y∞

1

1− z a_n

! e^Q^ν^(z)

is absolutely and uniformly convergent for|z|<R, however large R may

28

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4. Entire Functions 27 be.

Choose an R > 1, and anαsuch that 0< α < 1; then there exists a q (+ve integer) such that

|aq| ≤ R

α, |a_q+1|> R α. Then the partial product

q

Y

1

1− z a_n

! e^Q^ν^(Z)

is trivially an entire function of z. Consider the remainder Y∞

q+1

1− z a_n

! e^Q^ν^(z) for|z| ≤R.

We have, for n > q, |a_n| > R α or

z a_n

< α < 1. Using this fact we shall estimate each of the factors{1+∪n(z)}in the product, for n>q.

| ∪n(z)|=

e⁻^ν+1¹

_z

an

ν+1

−...

−1

≤e

1 ν+1

_z

an

_ν+1 ...

−1

Since |e^m−1| ≤ e^|m|−1, for e^m−1 = m+ m²

2! +. . .or|e^m−1| ≤ ²⁹

|m|+ |m²|

2 . . .=e^|m|−1.

|Un(z)| ≤e

z an

ν+1 1+

z an

+...

−1,

≤e

z an

ν+1(1+α+α²+···)

−1

=e^1−α¹

z an

ν+1

−1

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28 4. Entire Functions

≤ 1

1−α z a_n

ν+1

·e^1−α¹

z an

ν+1

, since if x is

real, e^x−1≤ xe^x, for e^x−1= x 1+ x 2!+ x²

3! +· · ·

!

≤ x(1+x+ x² 2! +

· · ·)= xe^x Hence

|Un(z)| ≤ e^1−α¹ 1−α

z an

ν+1

Now arise two cases: (i) either there exists an integer p > 0 such

30

that P^∞

n=1

1

|a_n|^p < ∞or (ii) there does not. In case (i) takeν = p−1, so that

|U_n(z)| ≤ R^p

|a_n|^p · e¹⁻¹^a

1−α, since|z| ≤R;

and henceP

|U_n(z)|<∞for|z| ≤R, i.e. Q^∞

q+1

(1+U_n(z)) is absolutely and uniformly convergent for|z| ≤R.

In case (ii) takeν=n−1, so that

|U_n(z)|< e¹⁻¹^α 1−α

z a_n

n

,n>q z a_n

<1

|z| ≤R

|an| → ∞ Then by the ‘root-test’P

|U_n(z)| < ∞, and the same result follows as before. Hence the product

Y∞ 1

1− z a_n

! e^Q^ν^(z)

is analytic in|z| ≤R; since R is arbitrary andνdoes not depend on R we

31

see that the above product represents an entire function.

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4. Entire Functions 29 Remark. If in addition z= 0 is also a zero of G(z) then G(z)

z^mG(z) has no zeros and equals e^g(z), say, so that

G(z)=e^g(z)·z^m Y∞ n=1

1− z a_n

! e^Q^ν^(z)

In the above expression for G(z), the function g(z) is an arbitrary entire function. If G is subject to further restrictions it should be possible to say more about g(z); this we shall now proceed to do. The class of entire functions which we shall consider will be called “entire functions of finite order”.

Definition. An entire function f (z) is of finite order if there exists a con- stantλsuch that|f (z)| < e^r^λ for|z| = r > r₀. For a non-constant f , of finite order, we haveλ > 0. If the above inequality is true for a certain λ, it is also true forλ^′ > λ. Thus there are an infinity ofλ^′s o satisfying this. The lower bound of such λ^′s is called the “order of f ”. Let us denote it byρ. Then, givenε >0, there exists an rosuch that

|f (z)|<e^r^ρ+ε for|z|=r >r0.

This would imply that M(r)=max

|z|=r |f (z)|<e^r^ρ+ε,r >r₀,

while 32

M(r)>e^r^ρ⁻^ε for an infinity of values of r tending to+∞

Taking logs. twice, we get log log M(r)

log r < ρ+ε and

log log M(r)

log r > ρ−εfor an infinity of values of r→ ∞

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30 4. Entire Functions Hence

ρ= lim

r→∞

log log M(r) log R

Theorem 2. If f (z) is an entire function of orderρ < ∞, and has an infinity of zeros, f (0), 0, then givenε >0, there exists an R_◦such that for R≥Ro, we have

n R

3

≤ 1

log 2·log e^R^ρ+ε

|f (o)|

Here n(R) denotes the number of zeros of f (z) in|z| ≤R.

Proof. We first observe that if f (z) is analytic in|z| ≤ R, and a₁. . .a_n are the zeros of f inside|z|<R/3, then for

g(z)= f (z) 1− _a^z₁

. . . 1− _a^z_n we get the inequality

33

|g(z)| ≤ M

2n

where M is defined by:|f (z)| ≤ M for|z|=R. For, if|z|=R, then since

|ap| ≤ R

3, p=1. . .n, we get z ap

≥3 or

1− z ap

≥2 By the maximum modules theorem,

|g(o)| ≤ M

2n, i.e.|f (o)| ≤ M

2n or

n≡n R

3

≤ 1

log 2·log M f (o)

!

If, further, f is of orderρ, then for r >r_◦, we have M<e^r^ρ+ε which gives the required results.

N.B. The result is trivial if the number of zeros is finite.

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4. Entire Functions 31 Corollary.

n(R)=O(R^ρ+ε).

For

n R

3

≤ 1 log 2·

R^ρ+ε−log|f (o)| , and −log|f (0)|<R^ρ+ε, say.

Then for R≥R^′

n R

3

< 2 log 2R^ρ+ε and hence the result.

Theorem 3. If f (z) is of orderρ <∞, has an infinity of zeros 34

a₁,a₂, . . . , f (o),0, σ > ρ, then X 1

|a_n|^σ <∞ Proof. Arrange the zeros in a sequence:

|a1| ≤ |a2| ≤. . . Let

α_p=|a_p|

Then in the circle|z| ≤r =α_n, there are exactly n zeros. Hence n<cα^ρ+ε_n , for n≥ N

or 1

n > 1 c · 1

α^ρ+ε_n Letσ > ρ+ε(i.e. choose 0< ε < σ−ρ). Then

1

n^σ/ρ+ε > 1 c^σ/ρ+ε · 1

α^σ_n,

or 1

α^σ_n <c^σ/ρ+ε· 1

n^σ/ρ+ε for n≥N HenceP 1

α^σ_n <∞.

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32 4. Entire Functions Remark. (i) There cannot be too dense a distribution of zeros, since n<c< α^ρ+ε_n . Nor can their moduli increase too slowly, since, for instance,P 1

(log n)^p does not converge.

(ii) The result is of course trivial if there are only a finite number of

35

zeros.

Definition. The lower bound of the numbers ‘σ’ for whichP 1

|a_n|σ <∞, is called the exponent of convergence of{an}. We shall denote it byρ₁.

Then

X 1

|an|^ρ¹^+ε <∞, X 1

|an|^ρ¹⁻^ε =∞, ε >0 By Theorem 1, we haveρ₁≤ρ

N.B. If the a^′_ns are finite in number or nil, then ρ₁ = 0. Thusρ₁ > 0 implies that f has an infinity of zeros. Let f (z) be entire, of orderρ <∞; f (o) , 0, and f (zn) = 0, n = 1,2,3, . . .. Then there exists an integer (p+1) such thatP 1

|zn|^ρ+1 <∞

(By Theorem 1, any integer > ρ will serve for ρ+ 1). Thus, by Weierstrass’s theorem, we have

f (z)=e^Q(z) Y∞

1

1− z z_n

! e

z

zn+_2zn2^z2 +···+_νz^z^νν n,

whereν= p (cf. proof of Weierstrass’s theorem).

Definition. The smallest integer p for whichP ₁

|zn|^p+1 <∞is the ‘genus’

of the ‘canonical product’ Π 1− z zn

! e

z zn+···+^zp

pzp

n, with zeros z_n. If z^′_ns

36

are finite in number, we (including nil) define p= 0, and we define the product asΠ 1− z

zn

!

Examples. (i) zn=n, p=1 (ii) zn=eⁿ, p=0 (iii) z₁= ¹₂log 2, z_n=log n, n≥2, no finite p.

Lectures on The Riemann Zeta–Function

Lectures on

The Riemann Zeta–Function

Lectures on the Riemann Zeta-Function

Contents

Lecture 1

The Maximum Principle

Lecture 2

The Phragmen-Lindelof principle

Lecture 3

Schwarz’s Lemma

Lecture 4

Entire Functions